Math & Logic Puzzles - Problem Solving

The series corresponds to the squares of integers: A=1 (1^2), D=4 (2^2), I=9 (3^2), P=16 (4^2). The next term is 5^2 = 25. The 25th letter of the alphabet is Y.

The pattern is n^2 + n starting from n=1: 1^2+1=2, 3^2+3=12, 5^2+5=30, 7^2+7=56. The next term uses the next odd number n=9: 9^2+9 = 81+9 = 90.

If he sells 40%, he has 60% left. 60% of Total = 300. Total = 300 / 0.60 = 500. Since 500 is not listed in the options (750, 450, 860), 'None of these' is the correct choice.

Cost price of 1 toffee = 1/6 rupee. For 20% profit, selling price = 1/6 * 1.2 = 1/5 rupee. Therefore 5 toffees must be sold for 1 rupee.

A = x% of y = (x/100) * y = xy/100. B = y% of x = (y/100) * x = xy/100. Since both expressions equal xy/100, A = B.

Train speed = 54 km/hr = 15 m/s. In 20 seconds (passing man), train length = 15 * 20 = 300 m. In 36 seconds (passing platform), distance covered = 15 * 36 = 540 m. Platform length = 540 - 300 = 240 m.

Let Johnny's age = x. Father = 4x (three times more means four times as much). After 8 years: 4x + 8 = 2.5(x + 8). Solving: 4x + 8 = 2.5x + 20, so 1.5x = 12, x = 8. After 16 more years (total 24 years from now), Johnny = 32, Father = 64. Ratio = 64/32 = 2 times.

Let Rita's age = x. Geeta = 2x (twice as old as Rita). Seeta = 2x + 4 (four years older than Geeta). Total: x + 2x + 2x + 4 = 54. So 5x = 50, x = 10. Geeta's age = 2x = 20 years.

Runs scored in first 10 overs = 3.2 * 10 = 32. Remaining runs needed = 282 - 32 = 250. Required run rate = 250/40 = 6.25 runs per over.

To have an average of zero, the sum of the 20 numbers must be zero. To maximize the count of positive numbers, 19 numbers can be positive (e.g., all 1s), and the 20th number can be a large negative number (e.g., -19) to balance the sum to zero.

Let marks be x and x+30. Higher score = 60% of sum, so x+30 = 0.60(2x+30). Solving: x+30 = 1.2x + 18, so 0.2x = 12, x = 60. The marks are 60 and 90.

The trains approach each other at 40+60=100 kmph, so they meet in 5 hours (500/100). The bird flies for 4 hours (one hour before meeting), covering 10×4=40 km. Option C is a distractor invoking infinite series unnecessarily - this is a simple time×speed problem. Option D is a joke option.

This is a perimeter counting problem. With 27 nails on each side: 4 × 27 = 108. However, each corner nail is counted twice (once for each adjacent side), so we subtract the 4 corner nails: 108 - 4 = 104 nails total.

If 8 monkeys eat 8 bananas in 8 minutes, it means each monkey takes 8 minutes to eat 1 banana (they are eating simultaneously). Therefore, 3 monkeys eating 3 bananas will also take 8 minutes, as each monkey still takes 8 minutes for its own banana.

Relative speed = 9 + 2 = 11 mph. Time to meet = Distance / Relative Speed = 33 / 11 = 3 hours. In 3 hours, the hare travels 9 mph * 3 hours = 27 miles.

Let B be bicycles and T be tricycles. B + T = 12 and 2B + 3T = 30. From the first eq, B = 12 - T. Substitute: 2(12 - T) + 3T = 30 => 24 - 2T + 3T = 30 => T = 6. There are 6 tricycles.

Since the money doubles every day and the piggy bank fills up completely on the 13th day, it means that on the preceding day, which is the 12th day, the piggy bank was exactly half as full. Doubling that half-full amount on day 13 fills it completely.

Let total life = x years. As boy: x/4, as youth: x/8, as active man: x/2, as old man: 12 years. Total: x/4 + x/8 + x/2 + 12 = x. Converting to eighths: 2x/8 + x/8 + 4x/8 + 12 = x, so 7x/8 + 12 = x, which means 12 = x/8, so x = 96 years. Active man period = x/2 = 48 years.

To find the sum of the numbers from 1 to 36, there is a formula that can be used. This formula is known as the arithmetic series formula:

$$S = \frac{n}{2}(a + l)$$

Where:

• S is the sum of the series
• n is the number of terms in the series
• a is the first term of the series
• l is the last term of the series

In this case, we have:

• n = 36 (as we are adding the numbers from 1 to 36)
• a = 1 (the first term of the series)
• l = 36 (the last term of the series)

Substituting these values into the formula:

$$S = \frac{36}{2}(1 + 36)$$

$$S = 18(37)$$

$$S = 666$$

Hence, the sum of the numbers from 1 to 36 is 666.

Therefore, the correct answer is: C. 666

The snail gains 4 ft net each day (6 up, 2 down). On the last day, it will reach the top during its 6 ft climb and not slip back. 74 - 6 = 68 ft. It takes 68/4 = 17 days to reach 68 ft. On the 18th day, it climbs 6 ft to reach 74 ft.